Here are the steps to solve these problems:
1. The points (x1, y1) and (x2, y2) form a diameter of a circle. The point (x, y) is another point on the circle.
(a) The gradient of the diameter AB is (y2-y1)/(x2-x1).
(b) The equation of AB is y-y1 = (y2-y1)/(x2-x1)(x-x1)
(c) Since P lies on AB, substitute the point (x, y) into the equation of AB to determine the value of x.
2. A line with equation y=mx+